surveying iindl.ethernet.edu.et/bitstream/123456789/87883/4/chapter 3...engineering surveying curve...
TRANSCRIPT
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Surveying II CENG 2092
SCHOOL OF CIVIL AND
ENVIROMENTAL ENGINEERING
Chapter 3
Curves
Tamru T.
2012EC (2019/20GC)
2nd Sem
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Engineering Surveying -- Def The ASCE defines engineering surveying as those activities involved in the planning and execution of surveys for the location, design, construction, operation, and maintenance of civil and other engineered projects. Such activities include:
The preparation of survey and related mapping specifications;
Execution of photogrammetric and field surveys for the collection of required data
Calculation, reduction and plotting of survey data for use in engineering design including use for geographic information systems (GIS);
Design and provision of horizontal and vertical control survey networks;
Monitoring of ground and structural stability, including alignment observations, settlement levels, and related reports and certifications; and
Analysis of errors and tolerances associated with the measurement, field layout and mapping.
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Engineering surveying
Curve setting out
Earth work computation
Staking of structures like building, bridge,
culvert and tunnel
Monitoring of ground and structural
stability
Route surveying
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Measurement and setting out
Accurate large-scale plan planning and
design of a construction project
The plan set out on the ground in the
correct absolute and relative position and to
its correct dimensions
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Curve
The center line of a road consists of series of
straight lines interconnected by curves that are
used to change the alignment, direction, or
slope of the road.
Those curves that change the alignment or
direction are known as horizontal curves, and
Those that change the slope are vertical
curves.
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Curve Setting out
Horizontal Curve Circular Curve
Compound Curve
Reverse Curve
Transition Curve
Vertical Curve Parabolic curve
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Simple circular and Compound
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Reverse Curve and Transition
curve
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Vertical Curves
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Horizontal Curves
The principal consideration
in the design of a curve is
the selection of the length
of the radius or the degree
of curvature
This selection is based on
such considerations as the
design speed of the
highway and the sight
distance as limited by
headlights or obstructions
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Elements of Horizontal Curves
I – Point of intersection
T1 – Point of
commencement
T2 – Point of tangency
E – External distance
T1T2 – Length of the
curve
T1I = IT2 = tangent
length
AI – Back Tangent
BI – Forward Tangent
T1T2 – Chord length
- Deflection angle
R – Radius of the curve
- Angle of intersection
M – Mid-ordinate
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Formula to calculate the various elements of
a circular curve for use in design and setting
out
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Setting out of circular curves
Offset from long chord
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Offset from tangent
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Setting out using one theodolite and tape
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Setting out using two theodolite
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Setting out using Coordinates
• Coordinates of T1 and I known form design
• known from deflection angle method
• Bearing of T1I is computed
• Using s and bearing of T1I bearings of T1A, T1B ..
• Distance T1A, T1B .. can be computed as 2Rsin(1) …
• using the bearing and distances coordinates of A, B, C .. are obtained
• These points can now be set out from the nearest control points by polar or intersection
methods
• By computing bearing and distance
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Setting out with inaccessible intersection
point
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Example 1
A circular curve of 500 m radius to be set out joining the two straights with deflection angle of 38. Calculate the necessary data for setting out the curve by the method of
Offset from long chord
Offset from tangent
Deflection angle method
Take peg interval of 30 m length and the chainage of I (60 + 13.385)
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Solution
Calculations of elements of the Curve
Tangent length = Rtan(/2) = 500tan19 = 172.164 m
Length of the curve (l) = (R/180) = 331.613 m
Chainage of T1 = Chainage of I – T1I = 60x30 + 13.385 – 172. 164
= 1641.221 = 54 + 21.22
Chainage of T2 = Chainage of I + l = 1641.221 + 331.613
= 1972.834 = 65 + 22.83
Long chord (L)= 2Rsin(/2) = 325.57 m
L/2 = 162.785 m
To locate the points on the curve, for distances
X = 0, 30, 60, 90, 120, 150, 162.785 m
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The offset from long chord is calculated as:
The calculated value of y,
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Offset from tangent
x 0 (T1) 30 60 90 120 150 172.164 (I)
y 0 0.901 3.613 8.167 14.614 23.030 30.575
Deflection angle method
Length of first sub-chord = (54 + 30) – (54 + 21.22) = 8.78 m
Length of last sub-chord = (65 + 22.83) – (65 + 0) = 22.83 m
Number of normal chords N = 65 – 55 = 10
Total number of chords n = 10 + 2 = 12
Tangential angle for the first chord = 1718.9*8.78/500 = 30.184 min
Tangential angle for the normal chord = 1718.9*30/500 = 103.184 min
Tangential angle for the last chord = 1718.9*22.83/500 = 78.485 min
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Point
Chainage
Chord
Length (m)
Tangential
angle (‘)
Deflection
Angle (‘)
Angle set on 1’’
0(T1) 54 + 21.22 - 0.0 0.0
1 55 + 00 8.78 30.184 30.184 0030’11’’
2 56 + 00 30 103.134 133.318 0213’19’’
3 57 + 00 30 103.134 236.452 0356’27’’
4 58 + 00 30 103.134 339.586 0539’35’
5 59 + 00 30 103.134 442.720 0722’43’’
6 60 + 00 30 103.134 546.104 0906’06’’
7 61 + 00 30 103.134 648.988 1048’59’’
8 62 + 00 30 103.134 752.122 1232’07’’
9 63 + 00 30 103.134 855.256 1415’15’’
10 64 + 00 30 103.134 958.839 1558’50’’
11 65 + 00 30 103.134 1061.524 1741’31’’
12(T2) 65 + 22.83 22.83 78.485 1140.009 1900’01’’
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Transition curves
The transition curve is a curve of constantly changing radius
Some of the important properties of the spirals are given
below:
•L = 2Rθ
•θ = (L / Ls)2 * θs
•θs = Ls / 2Rc (in radians) = 28.65Ls / Rc (in degrees)
•Ts = Ls /2 + (Rc + S)*tan(Δ/2)
•S = Ls2 / 24Rc
•Es = (Rc + S)*sec(Δ/2) - Rc
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Transition curves
Note:
θs = spiral angle
Δ = total central angle
Δc = central angle of the circular arc extending fro BC to EC = Δ - 2
θs
Rc = radius of circular curve
L = length of spiral from starting point to any point
R = radius of curvature of the spiral at a point L distant from
starting point.
Ts = tangent distance
Es = external distance
S = shift
HIP = horizontal intersection point
BS = beginning of spiral
BC = beginning of circular curve
EC = end of circular curve
ES = end of spiral curve
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Shift, S
Where transition curves are introduced
between the tangents and a circular curve of
radius R, the circular curve is “shifted”
inwards from its original position by an
amount BP = S so that the curves can meet
tangentially.
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Setting out of transition curve
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Example 2
Two straights AB and BC intersect at chainage 1530.685 m, the total deflection angle being 33°08′. It is proposed to insert a circular curve of 1000 m radius and the transition curves for a rate of change of radial acceleration of 0.3 m/s3, and a velocity of 108 km/h. Determine setting out data using theodolite and tape for the transition curve at 20 m intervals and the circular curve at 50 m intervals.
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Solution = 0.3 m/cub sec
V = 108 km/hr = 30 m/s
= 3308’ /2 = 16 34’
Radius of circular curve = 1000 m
Chainage of I = 1530.686 m
Peg interval for transition curve = 20 m
Peg interval for circular curve = 50 m
m
xR
LSShift
mxR
VL
LR
V
338.0100024
90
24,
0.9010003.0
30
22
333
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Tangent length IT= (R+S)tan(/2) + (L/2)
= 342.580
1= L/2R rad = 90/(2x1000) x(180/)= 234’42’’
Angle subtended by T1- T2 = = - 21 = 2758’36’’
Length of the curve, l = (R/180) = 488.285 m
Chainage of T = Chainage of I – IT = 1188.105 m
= 59 + 8.105 chains for peg interval 20 m
Chainage of T1 = Chainage of T + L = 1278,105 m
= 25 + 28.105 chains for peg interval 50 m
Chainage of T2 = Chainage of T1 + l = 1766.390 m
= 35 + 16.390 chains for peg interval 50 m
= 88 + 6.390 chains for peg interval 20 m
Chainage of U = Chainage of T2 + L = 1856.390 m
= 92 + 16.390 chains for peg interval 20 m
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Length of the first sub chord for transition
(59 + 20) – (59 + 8.105) = 11.895 m
Length of first sub chord for circular curve
(25 + 50) – (25 + 28.105) = 21.895 m
Length of last sub chord for circular curve
(35 + 16.390) – (35 + 0) = 16.390 m
Deflection angle for transition curve
Deflection angle for circular curve
min006366.01800 2
2
lRL
l
min9.1718R
c
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Setting out data for 1st transition curve
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Setting out data for circular curve
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Setting out data for 2nd Transition curve
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Vertical Curve
They should be of sufficiently large curvature
to provide comfort to the driver, that is, they should
have a low ‘rate of change of grade’.
Vertical curves (VC) are
used to connect intersecting
gradients in the vertical
plane.
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Vertical curve formula
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Properties of a Vertical Curve The difference in elevation between the BVC and
a point on the g1 grade line at a distance X units
(feet or meters) is g1X (g1 is expressed as a
decimal).
The tangent offset between the grade line and the
curve is given by ax2, where x is the horizontal
distance from the BVC; (that is, tangent offsets are
proportional to the squares of the horizontal
distances).
The elevation of the curve at distance X from the
BVC is given (on a crest curve) by:
BVC + g1x - ax2
(the signs would be reversed in a sag curve).
The grade lines (g1 and g2) intersect midway
between the BVC and the EVC. That is, BVC to V
= 1/2L = V to EVC.
Offsets from the two grade lines are symmetrical
with respect to the PVI.
The curve lies midway between the PVI and the
midpoint of the chord; that is, Cm = mV.
L L
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Computation of Low or High
Point on Curve The locations of curve high and low points are
important for drainage and bridge considerations. For
example, on curbed streets catch basins must be
installed precisely at the drainage low point.
It was noted earlier that the slope was given by the
expression
Slope = 2ax + g1
The figure above shows a sag vertical curve with a
tangent drawn through the low point; it is obvious that
the tangent line is horizontal with a slope of zero; that
is,
2ax + g1 = 0
Since 2a = A/L
x = -g1L/A
where x is the distance from the BVC to the high or
low point.
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Procedure for Computing a
Vertical Curve
1. Compute the algebraic difference in grades: A
2. Compute the chainage of the BVC and EVC. If the
chainage of the PVI is known,
1/2L is simply subtracted and added to the PVI
chainage.
3. Compute the distance from the BVC to the high or
low point (if applicable):
x = -g1L/A
and determine the station of the high/low point.
4. Compute the tangent grade line elevation of the BVC
and the EVC.
5. Compute the tangent grade line elevation for each
required station.
6. Compute the midpoint of chord elevation
{elevation of BVC + elevation of EVC}/2
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Procedure for Computing a
Vertical Curve 7. Compute the tangent offset (d) at the PVI (i.e.,
distance Vm):
d = {elevation of PVI - elevation of midpoint of chord}/2
8. Compute the tangent offset for each individual station.
Tangent offset = {x/(L/2)}2d
where x is the distance from the BVC or EVC (whichever is
closer) to the required station.
9. Compute the elevation on the curve at each required
station by combining the tangent offsets with the
appropriate tangent grade line elevations. Add for
sag curves and subtract for crest curves.
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Example 3
A vertical curve 120 m long of the parabola type is
to join a falling gradient of 1 in 200 to a rising
gradient of 1 in 300. If the level of the intersection
of the two gradients is 30.36 m. Compute the
levels at 15-m intervals along the curve.
1/200
(p = -0.5%)
1/300
q = (+0.33%)
30.36
Chainage 2000
B
A
C
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Solution
Chainage of A = Chainage of B – L/2
= 2000 – 60 = 1940 m
Chainage of C = Chainage of B + L/2
= 2000 + 60 = 2060 m
Elevation of A = Elev. B + (pL/200)
= 30.36 + (0.5*120/200) =30.660
Elevation of C = Elev. B + (qL/200)
= 30.36 + (0.33*120/200) = 30.56 m
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Station Chainage (m) x(m) yi
A (BVC) 1940 0 30.660
1 1955 15 30.593
2 1970 30 30.541
3 1985 45 30.505
4 (above B) 2000 60 30.485
5 2015 75 30.479
6 2030 90 30.490
7 2045 105 30.516
C ( EVC) 2060 120 30.558 30.56
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Example 4
Given that L = 300 m, g1= -3.2%, g2= +1.8%, PVI at
30 + 030, and elevation = 465.92. Determine the
location and elevation of the low point and elevations
on the curve at 50m interval starting from BVC.
1 st =100